# Find the arc length of the cuve r(t)=(cos^3 t) i +  (sin^3 t) j +  2k     ,     0=< t =<phi/2 solve this early You need to determine the length of the vector function `bar r(t) =`  `ltcos^3 t, sin^3 t, 2gt, ` over interval `[0,pi/2], ` hence you need to find the magnitude of the tangent vector such that:

`|bar r'(t)| = sqrt((3cos^2 t*(-sin t))^2 + (3sin^2 t*cos t)^2)`

`|bar r'(t)| = sqrt(9cos^4...

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You need to determine the length of the vector function `bar r(t) =`  `ltcos^3 t, sin^3 t, 2gt, ` over interval `[0,pi/2], ` hence you need to find the magnitude of the tangent vector such that:

`|bar r'(t)| = sqrt((3cos^2 t*(-sin t))^2 + (3sin^2 t*cos t)^2)`

`|bar r'(t)| = sqrt(9cos^4 t*sin^2 t + 9sin^4 t*cos^2 t) `

`|bar r'(t)| = sqrt(9cos^2 t*sin^2 t(cos^2 t+ sin^2 t) `

You need to remember that `cos^2 t + sin^2 t = 1`  such that:

`|bar r'(t)| = sqrt(9cos^2 t*sin^2 t)`

`|bar r'(t)| = 3 sin t*cos t`

You need to remember the formula of arc length of a curve such that:

`L = int_0^(pi/2) |bar r'(t)| dt`

`L = int_0^(pi/2) 3 sin t*cos t dt`

You need to multiply and divide the integrand `sin t*cos t`  by 2 to obtain the formula of sine of double angle such that:

`L = (3/2) int_0^(pi/2)2 sin t*cos t dt`

`L = (3/2) int_0^(pi/2) sin(2t) dt `

`L = -(3/2) (cos(2t))/2 |_0^(pi/2)`

`L = -(3/4)(cos pi - cos 0)`

`L = -(3/4)(-1- 1)`

`L = 3/2`

Hence, evaluating the arc length of the curve given in vectorial form yields `L = 3/2` .

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